私はphpとmysqlで自分のサイトのページネーションコードをコーディングしましたが、レコードがないmysql_fetch_array() expects parameter 1 to be resource
場合、このコードではデータベーステーブルが空であるPage 0 of 0 Back -1 0
場合にエラーが発生し、recoredがデータベーステーブルにある場合は非常に正常に動作します. 私のphpページネーションコードは次のとおりです
$sql = mysql_query("SELECT id, firstname, country FROM myTable ORDER BY id ASC");
$nr = mysql_num_rows($sql);
if (isset($_GET['pn'])) {
$pn = preg_replace('#[^0-9]#i', '', $_GET['pn']);
} else {
$pn = 1;
}
$itemsPerPage = 10;
$lastPage = ceil($nr / $itemsPerPage);
if ($pn < 1) { // If it is less than 1
$pn = 1; // force if to be 1
} else if ($pn > $lastPage) { // if it is greater than $lastpage
$pn = $lastPage; // force it to be $lastpage's value
}
$centerPages = "";
$sub1 = $pn - 1;
$sub2 = $pn - 2;
$add1 = $pn + 1;
$add2 = $pn + 2;
if ($pn == 1) {
$centerPages .= ' <span class="pagNumActive">' . $pn . '</span> ';
$centerPages .= ' <a href="' . $_SERVER['PHP_SELF'] . '?pn=' . $add1 . '">' . $add1 . '</a> ';
} else if ($pn == $lastPage) {
$centerPages .= ' <a href="' . $_SERVER['PHP_SELF'] . '?pn=' . $sub1 . '">' . $sub1 . '</a> ';
$centerPages .= ' <span class="pagNumActive">' . $pn . '</span> ';
} else if ($pn > 2 && $pn < ($lastPage - 1)) {
$centerPages .= ' <a href="' . $_SERVER['PHP_SELF'] . '?pn=' . $sub2 . '">' . $sub2 . '</a> ';
$centerPages .= ' <a href="' . $_SERVER['PHP_SELF'] . '?pn=' . $sub1 . '">' . $sub1 . '</a> ';
$centerPages .= ' <span class="pagNumActive">' . $pn . '</span> ';
$centerPages .= ' <a href="' . $_SERVER['PHP_SELF'] . '?pn=' . $add1 . '">' . $add1 . '</a> ';
$centerPages .= ' <a href="' . $_SERVER['PHP_SELF'] . '?pn=' . $add2 . '">' . $add2 . '</a> ';
} else if ($pn > 1 && $pn < $lastPage) {
$centerPages .= ' <a href="' . $_SERVER['PHP_SELF'] . '?pn=' . $sub1 . '">' . $sub1 . '</a> ';
$centerPages .= ' <span class="pagNumActive">' . $pn . '</span> ';
$centerPages .= ' <a href="' . $_SERVER['PHP_SELF'] . '?pn=' . $add1 . '">' . $add1 . '</a> ';
}
$limit = 'LIMIT ' .($pn - 1) * $itemsPerPage .',' .$itemsPerPage;
$sql2 = mysql_query("SELECT id, firstname, country FROM myTable ORDER BY id ASC $limit");
if ($lastPage != "1"){
// This shows the user what page they are on, and the total number of pages
$paginationDisplay .= 'Page <strong>' . $pn . '</strong> of ' . $lastPage. ' ';
// If we are not on page 1 we can place the Back button
if ($pn != 1) {
$previous = $pn - 1;
$paginationDisplay .= ' <a href="' . $_SERVER['PHP_SELF'] . '?pn=' . $previous . '"> Back</a> ';
}
// Lay in the clickable numbers display here between the Back and Next links
$paginationDisplay .= '<span class="paginationNumbers">' . $centerPages . '</span>';
// If we are not on the very last page we can place the Next button
if ($pn != $lastPage) {
$nextPage = $pn + 1;
$paginationDisplay .= ' <a href="' . $_SERVER['PHP_SELF'] . '?pn=' . $nextPage . '"> Next</a> ';
}
}
$outputList = '';
while($row = mysql_fetch_array($sql2)){
$id = $row["id"];
$firstname = $row["firstname"];
$country = $row["country"];
$outputList .= '<h1>' . $firstname . '</h1><h2>' . $country . ' </h2><hr />';
}
<h2>Total Items: <?php echo $nr; ?></h2>
<div style="margin-left:58px; margin-right:58px; padding:6px; background-color:#FFF; border:#999 1px solid;"><?php echo $paginationDisplay; ?></div>
<div style="margin-left:64px; margin-right:64px;"><?php print "$outputList"; ?></div>
<div style="margin-left:58px; margin-right:58px; padding:6px; background-color:#FFF; border:#999 1px solid;"><?php echo $paginationDisplay; ?></div>
オプションを使用して適切なページネーションを提案できpage 1 of 1 from 1
ますか?